Problem Description

In this problem, you are given an array of integers nums and an integer goal. The objective is to select a subsequence from the array such that the absolute difference between the sum of the selected subsequence and the goal is minimized. In other words, you want to find a subsequence whose sum is as close as possible to the goal. A subsequence of an array can be derived by omitting any number of elements from the array, which could potentially be none or all of them.

Intuition

The direct approach to solve this problem would involve generating all possible subsequences of the given array nums and calculating the sum for each subsequence to see how close it is to the goal. However, this approach is not feasible because the number of possible subsequences of an array is 2^n, which would result in exponential time complexity making it impractical for large arrays.

The intuition behind the provided solution is to use the "meet in the middle" strategy. This involves dividing the array nums into two nearly equal halves and then separately generating all possible sums of subsequences for each half. Once you have all subset sums from both halves, you can sort one of the halves (for example, the right half) to then use binary search to quickly find the best match for each subset sum from the other half (the left half). This way, you have reduced the original problem, which would have a complexity of O(2^n), into two subproblems each having a complexity of O(2^(n/2)), which is significantly more manageable.

The minAbsDifference function first generates all possible subset sums for both halves of the array using the helper function getSubSeqSum. It stores these sums in two sets, left and right. After that, it sorts the sums from the right half to allow efficient searching. For each sum in the left set, the function computes the complement value needed to reach the goal. Using binary search, it then checks whether a sum in the right set exists that is close to this complement value. The result is the smallest absolute difference found during this pairing process between left and right subset sums.

Learn more about Two Pointers, Dynamic Programming and Bitmask patterns.

Solution Approach

The solution approach consists of the following key parts:

1. Divide the array into two halves: First, the original array nums is split into two halves. This is a crucial step in the "meet in the middle" strategy.

2. Generate all possible sums of both halves: The method getSubSeqSum is recursively called to calculate all feasible subset sums for the two halves of the array, which are stored in the left and right sets. This is done through classic backtracking – for each element, we can either choose to include it in the current subset or not, leading to two recursive calls.

   The pseudo-code for subset sum generation would be similar to:

   function getSubSeqSum(index, currentSum, array, resultSet):
       if index == length of array:
           add currentSum to resultSet
           return
       end if
   
       // Don't include the current element
       call getSubSeqSum(index + 1, currentSum, array, resultSet)
   
       // Include the current element
       call getSubSeqSum(index + 1, currentSum + array[index], array, resultSet)
   end function

3. Sort the sums of one half and use binary search: After the generation of all subset sums for both halves, the sums from the right half are sorted to leverage binary search. The binary search allows us to find the closest element in right to the goal - left_sum, giving us the potential minimum difference.

   The logic for binary search comparison is as follows:

   for each sum in left set:
       compute remaining = goal - sum
       find the index of the minimum element that is greater than or equal to remaining in the right
       if such an element exists, update result to minimum of result or absolute difference between that element and remaining
       if there is an element less than remaining (index > 0), update result to minimum of result or absolute difference between that element and remaining
   end for

4. Calculate and return the result: The overall minimum difference is tracked in the result variable, which is initialized with infinity (inf). It gets updated whenever a smaller absolute difference is found between a pair of left and right subset sums with respect to the goal.

By exploiting the "meet in the middle" strategy and binary search, we manage to reduce the time complexity of the problem from exponential to O(n * 2^(n/2)), which is much more efficient for inputs within the problem's constraints.

Example Walkthrough

Let's assume we have the following input:

nums = [1, 2, 3, 4] and goal = 6.

Following the solution approach described:

1. Divide the array into two halves: We split nums into left_half = [1, 2] and right_half = [3, 4].

2. Generate all possible sums of both halves: We use the getSubSeqSum method.

   For the left_half:

   Using index 0, with currentSum 0 -> left {0}
   Include number at index 0 -> left {0, 1}
   Using index 1, with currentSum 1 -> left {0, 1}
   Include number at index 1 -> left {0, 1, 3}
   Repeat without including number 1 -> left {0, 1, 3, 2}
   Include number at index 1 -> left {0, 1, 3, 2, 3}

   So the possible sums for left_half are {0, 1, 2, 3}.

   For the right_half:

   Using index 0, with currentSum 0 -> right {0}
   Include number at index 0 -> right {0, 3}
   Using index 1, with currentSum 3 -> right {0, 3}
   Include number at index 1 -> right {0, 3, 7}
   Repeat without including number 3 -> right {0, 3, 7, 4}
   Include number at index 1 -> right {0, 3, 7, 4, 7}

   So the possible sums for right_half are {0, 3, 4, 7}.

3. Sort the sums of one half and use binary search: We sort the right_half sums to {0, 3, 4, 7} (although it's already sorted in this example).

   We perform binary searches for complement values (goal - left_sum) for each sum in the left_half sums.

   left_sum = 0, remaining = goal - 0 = 6, closest in right is 7 (absolute difference 1)
   left_sum = 1, remaining = goal - 1 = 5, closest in right is 4 or 7 (absolute differences 1 and 2)
   left_sum = 2, remaining = goal - 2 = 4, closest in right is 4 (absolute difference 0)
   left_sum = 3, remaining = goal - 3 = 3, closest in right is 3 (absolute difference 0)

4. Calculate and return the result: From the binary search steps, we find that the smallest absolute difference is 0, which can be achieved with left_sum of 2 and right_sum of 4 or left_sum of 3 and right_sum of 3. Hence the minimum absolute difference between any subsequence sum and the goal is 0.

By using the described strategy, the problem that had a potentially exponential complexity is tackled in a much more manageable way, making it possible to solve efficiently even with larger inputs.

Solution Implementation

Time and Space Complexity

The time complexity and space complexity analysis for the minAbsDifference function is as follows:

Time Complexity:

• The function getSubSeqSum generates all possible subsets' sums for the input subarrays. This function is called recursively for each element in the input subarray. There are 2^n subsets possible for an array with n elements, resulting in a time complexity of O(2^n) for creating all subsets for an array.
• Since getSubSeqSum is first called with half the size of the input array O(2^(n/2)), and then with the remaining half O(2^(n/2)), the total time for the subset sum generation steps for both halves is O(2*(2^(n/2))) which simplifies to O(2^(n/2 + 1)).
• The set right is sorted afterwards which has at most 2^(n/2) elements leading to a sort time complexity of O(2^(n/2) * log(2^(n/2))) which simplifies to O((n/2) * 2^(n/2)).
• Next, the for loop iterates over all elements of left, and for each element, a binary search is performed on right using bisect_left. This gives a time complexity of O(2^(n/2) * log(2^(n/2))) for the loop which simplifies to O((n/2) * 2^(n/2)).

The overall time complexity can be expressed as O(2^(n/2 + 1) + (n/2) * 2^(n/2) + (n/2) * 2^(n/2)) which simplifies to O(n * 2^(n/2)).

Space Complexity:

• The space required is to store all subset sums for both halves and considering the deepest recursion call stack. This leads to O(2^(n/2)) for left and O(2^(n/2)) for right sets separately.
• The recursion call stack of the function getSubSeqSum will go to a maximum depth of n/2 in both halves, so the space used by the call stack is O(n/2 + n/2) which simplifies to O(n).

Hence, the space complexity can be viewed as the maximum space consumed at any point, giving O(2^(n/2) + 2^(n/2) + n) which simplifies to O(2^(n/2) + n).

Learn more about how to find time and space complexity quickly using problem constraints.